1200 16. QUASITHIN GROUPS OF EVEN TYPE BUT NOT EVEN CHARACTERISTICTherefore L* is M 12 , J 2 , HS, or Ru. If Z(L) -=/:-1, then from (5b) and (7b)
of I.2.2, the projection v of u on L is of order 4, so u = vt with t E Tc of order
4, contrary to <I>(Tc) = 1. Hence Z(L) = 1, so if v is the projection of u on Land there is l E L with vv^1 an involution of X, then by 16.5.10.6, vv^1 is not a
2-central involution of L'. However X ~As, As, A 6 , or Sz(8), respectively, withall involutions in X 2-central in L, and the involutions in vX are in vL, so we have
a contradiction which completes the proof of 16.5.11. DLEMMA 16.5.12. L* is of Lie type in characteristic 2 of Lie rank 2.
PROOF. Assume otherwise. By 16.5.11 and 16.5.10.1, Lis Ln(2) for n := 4 or
- Thus H is either L or Aut(L), so H = LKT. Recall z is an involution in
Tc n Z(T). If Tc is cyclic, then by 16.4.4.2, Tc= CK(z), so that Gz =LT.
Next L has two classes j1 and j 2 of involutions, where j 1 is the class of transvec-
tions and the 2-central class. Hence u E j 2 by 16.5.10.5, so by 16.1.4.3, X ~ A4
or Z 3 /2^4 +4, for n = 4 or 5, respectively. Also CT·(X) ~ E4, unless LT* ~ Ss,
in which case CT (X) ~ D 8. Thus by 16.5.6.2, either R is a subgroup of E 4 , or
H = L R ~ Ss and R ~ Z4; R is not Ds as D1(R) = U*::::; T[,.
Suppose H* ~ 88 with R ~ Z 4. Then Tc ~ R ~ Z4, so Gz = LRTc by
paragraph one. Hence T = TLTcR centralizes Tc, so Tc ::::; Z(Gz)· Then R ::::;Z(Ca(u)), whereas R ~ Z4 is not central in a subgroup D 8 of CL(u*).
Therefore R ~ Z2 or E4, so that R = U and hence R*::::; TI,. Let v denote the
projection of u on L.Assume first that n = 5. Then
(02(X)) =: V = [V, X] EB Cv(X),
with the involutions in the 4-groups [V, X] and Cv(X) of type j 2 , and the diagonal
involutions of type j1. Then v E Cv(X), and there is l E L with v^1 E [V, X], so
that vv^1 is 2-central in L', contrary to 16.5.10.6.
Hence n = 4, so that L ~ L4(2). Assume R ~ E4. Then for r E R - (u),
the projection of r on Lis in 02 (CL(X)) - (v), so as Ca(u) ::::; H' by 16.4.2.5,
v E [r,CTL(u)]::::; K', sou= v EL. Similarly r EL, so that R = 02(CL(X)).
Then there is y EL of order 3 faithful on R. But as m 3 (NL(0 2 (X))) = 2 and G is
quasithin, K is a 3'-group, soy E 031 (H') = L'::::; Ca(R), a contradiction.
Therefore R = (u) is of order 2, so as Tc ~ R, Tc = (z) is of order 2,and Gz = LT by paragraph one. Set A := 02(CL(u))Tc. Then A ~ E32 and
A= J(CTLTo(u)). If H* ~ L4(2) then T =TL x (z), so that z rj. <I>(T) with TE
Sylz(Ca(z)). However z^0 nTL -=f. 0 by Thompson Transfer, and each involution a EL satisfies a E ( CL( a)) (cf. parts (1) and (3) of 16.1.4). Therefore LT~ 88 x Z2
with Ds x Es~ 02(CLT(u)) = J(02(CLT(u))). Choose g with TnH' = NT(K')::::;
Tg as in 16.4.2.4, so that R = Tf!:: by 16.4.11.3, and hence u = zg E zG n A - {z}
and A::::; Tg.Suppose first that A ::::; L'K'. Then A ::::; Tf,Tf!::, so A = J(C(TLTo)g(z)) =
02(CL1(z)) x R, and hence A plays the same role for the pairs L',Tc and L,R.
Next Anj2 is an orbit oflength 6 on A#nL under NH(A), with AutH(A) ~ ot(2).
Further if y E G such that zY E LK projects on a member of the class j 1 , then
KY E .6.o by 16.4.9.3, contrary to 16.5.10.5. Thus no member of zG n LK projects
on J1, so as NH(A) has two orbits of length 6 on the elements of A projecting on